SOLUTION: please help me solve this problem: A chemist has 6 liter of a 50% acid solution. How much water must be added to obtain a 20% acid solution.- i tryed with these two problems: 6+.50
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: please help me solve this problem: A chemist has 6 liter of a 50% acid solution. How much water must be added to obtain a 20% acid solution.- i tryed with these two problems: 6+.50
Log On
Question 617858: please help me solve this problem: A chemist has 6 liter of a 50% acid solution. How much water must be added to obtain a 20% acid solution.- i tryed with these two problems: 6+.50=y and 6(.50)+y= .20x but i don't get the right answer please help me try to figure out what i did wrong Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A chemist has 6 liter of a 50% acid solution. How much water must be added to obtain a 20% acid solution.
Equation:
water + water = water
0.50*6 + x = 0.80(6+x)
Multiply thru by 100 to get:
50*6 + 100x = 80*6 + 80x
20x = 30*6
x = (3/2)6
x = 9 liters (amt. of water to be added)
=====================================
Cheers,
Stan H.
=====================================